Algebra One Problem

Question

Bella has a stack of 41 bills (ones, fives, and tens) in an envelope. She has 3 less tens than fives and the quantity of ones doubles the number of fives. Write and solve an equation to determine how many of each bill type she has.

Answer 1

Let F= # of Fives, T= # of Tens, W = # of Ones

W+F+T = 41

T=F-3

W=2F

2F+F+F-3 = 41

4F = 44

F =11 Fives

T = 8 Tens

W =22 Ones

Answer 2

a = # of ones

b = # of fives

c = # of tens

41 = a + b + c

# of ones = 2 (# of fives) can be written as

a = 2b

# of tens = # of fives - 3 can be written as

c = b - 3

Substitute it into the equation (41 = a + b + c)

41 = 2b + b + b - 3

41 = 4b - 3

4b = 44

b = 11

Plug "b" back into the equations to find "a" and "c"

a = 2(11) = 22

c = 11 - 3 = 8

Therefore, there are 22 ones, 11 fives, and 8 tens.